Sign In
JEE MAIN - Mathematics Hindi (2002 - No. 14)
यदि $$y = {\left( {x + \sqrt {1 + {x^2}} } \right)^n},$$ तो $$\left( {1 + {x^2}} \right){{{d^2}y} \over {d{x^2}}} + x{{dy} \over {dx}}$$ है
$${n^2}y$$
$$-{n^2}y$$
$$-y$$
$$2{x^2}y$$
Comments (0)
Login To Comment
Advertisement
Allow javascript to properly load this page